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Two poles are $10 m$ and $20 m$ high. The line joining their tops makes an angle of $15^{\circ}$ with the horizontal. The distance between the poles is approximately equal to

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36.3 m

37.3 m

38.3 m

39.3 m

answer is B.

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Detailed Solution

It is given that two poles are

10 m

and

20 m

in height and the line which joins their tops makes an angle of

15^{\circ}

with the horizontal.
According to the given data the figure has been drawn below.
Question Image

Let

AB

and

CD

are two poles.
Then,

AB = ED = 10 m

and

CD = 20 m

and

∠ CAE = 15^{\circ}

.
Here,

CE = CD - ED

\Rightarrow 20 - 10

\Rightarrow 10

m
In

△ ACE,

tan θ = P e r p e n d i c u l a r b a s e

\Rightarrow \tan 15 ° = \frac{CE}{AE}

\Rightarrow AE = \frac{10}{\tan 15 °}

Now

\tan 15 ° = \tan (45 ° - 30 °)

= \frac{\tan 45 ° - \tan 30 °}{1 + \tan 45 ° \tan 30 °}

∵ tan (a − b) = tan a − tan b 1 + tan a tan b

= \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}

∵ tan 30 ° = 13 and tan 45 ° = 1

= \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}}

= \frac{\sqrt{3} - 1}{\sqrt{3} + 1}

Rationalizing denominator we get,

\frac{\sqrt{3} - 1}{\sqrt{3} + 1}

\Rightarrow (\sqrt{3} - 1)^{2} \times \frac{1}{2}

\Rightarrow (4 - 2 \sqrt{3}) \times \frac{1}{2}

\Rightarrow (2 - \sqrt{3}

)
So the value of AE =

\frac{10}{2 - \sqrt{3}}

m.
Now rationalizing the denominator, we must multiply both denominator and numerator by 2+

\sqrt{3} .

We get

\frac{10 \times (2 + \sqrt{3})}{(2 - \sqrt{3}) \times (2 + \sqrt{3})}

\Rightarrow \frac{10 (2 + \sqrt{3})}{4 - 3}

\Rightarrow \frac{10 (2 + 1.732)}{1}

∵ 3 = 1.732

\Rightarrow 37.3

\Rightarrow AE = 37.3 m .

Therefore the distance between the two poles is

37.3 m

.
So the correct option is 2.

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